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(Cross-posted from https://math.stackexchange.com/questions/4931582/steinitz-isomorphism-theorem-for-non-dedekind-domains)

Fix a Dedekind domain $R$ and fractional ideals $I, J$. It's a classical result by Steinitz that $I\oplus J \cong R \oplus IJ$ as $R$-modules.

Question Does this still hold for non-Dedekind 1-dimensional Noetherian domains, esp. non-maximal orders in number fields? Does it hold when assuming that $I,J$ are invertible?

I've seen this stated without proof and can't seem to find good literature on the topic. Literature recommendations on extending basic results on Dedekind domains to arbitrary orders in number fields are welcome. For example, does the classification of f.g. projective modules as direct sums of ideals carry over from Dedekind domains to orders?

Once one reduces to $I,J$ integral and $I+J=R$, the result is completely general (see e.g. https://math.stackexchange.com/questions/3297962/let-r-be-a-ring-prove-i-oplus-j-cong-r-oplus-ij-as-r-modules?noredirect=1). However, I can't find a reduction to this case generalized beyond Dedekind domains.

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2 Answers 2

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Here is a general fact attributed to Serre. Let $A$ be a Noetherian ring of Krull dimension $d$. If $P$ is a projective module over $A$ of rank larger than $d$, then $P\cong Q\oplus A$. It is easy to deduce that if $d=1$ and $I$, $J$ are invertible, then $I\oplus J\cong A\oplus IJ$.

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  • $\begingroup$ Thank you,@Mohan. Can you give a reference? $\endgroup$ Commented Jun 14 at 18:35
  • $\begingroup$ @DrBrownBear Many books have it. For example, you can find it in Mumford's "Curves on Surfaces". $\endgroup$
    – Mohan
    Commented Jun 14 at 18:47
  • $\begingroup$ Does it go under a particular name? Would I usually find this in the above formulation, or a more geometric one in terms of vector bundles? I don't have Mumford's book, but a few others on algebraic geometry and commutative algebra. $\endgroup$ Commented Jun 14 at 20:39
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    $\begingroup$ For the record, see e.g. Theorem 2.3 in maths.ed.ac.uk/~v1ranick/papers/kbook.pdf . Thanks Mohan! $\endgroup$ Commented Jun 14 at 20:53
  • $\begingroup$ This result also affimatively answers the side-question I had: Every rank n projective module over an order isomorphic to a sum of ideals. $\endgroup$ Commented Jun 15 at 3:48
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We answer in the negative the question as to whether there is an $R$-module isomorphism $$I \oplus J \simeq R \oplus IJ$$ whenever $I$ and $J$ are non-invertible ideals of a non-maximal order $R$ of some quadratic number field $K$.

Let $$ F \xrightarrow[]{\varphi} G \rightarrow M \rightarrow 0 $$ be an exact sequence such that $F$ and $G$ are free modules over $R$ and $G$ is finitely generated. Let $I_i(\varphi)$ the ideal of $R$ generated by the $i \times i$ minors of the matrix of $\varphi: F \rightarrow G$, agreeing that $I_i(\varphi) = R$ if $i \le 0$.
Then the $(i + 1)$-th Fitting ideal of $M$ ($i \ge 0$) is defined as $$\text{Fitt}_i(M) := I_{\text{rank}(G) - i}(\varphi).$$

The Fitting ideals are independent of $\varphi$ by Fitting's lemma [1, Corollary 20.4], i.e., they are invariant under isomorphisms of $R$-modules.

Proposition. Let $I, J$ be two non-zero ideals of an integral domain $R$. Assume moreover that each of $I, J$ and $IJ$ is $2$-generated. Then we have

  • $\text{Fitt}_2(I \oplus J) = \text{Fitt}_1(I) \text{Fitt}_1(J)$ and $\text{Fitt}_3(I \oplus J) = \text{Fitt}_1(I) + \text{Fitt}_1(J)$

while

  • $\text{Fitt}_2(R \oplus IJ) = \text{Fitt}_1(IJ)$ and $\text{Fitt}_3(R \oplus IJ) = R$.

Proof. Observe that $\text{Fitt}_0(I) = \text{Fitt}_0(J) = \{0\}$ [1, Proposition 20.7.a] and $I_r(\varphi \oplus \psi) = \sum_{s + t = r} I_s(\varphi) I_t(\psi)$ (use for instance [1, Definition 20.2]).

Corollary. Let $I$ be a $2$-generated non-zero ideal of an integral domain $R$. Assume moreover that $I$ is not invertible and that $I^2 := I \cdot I$ is $2$-generated. Then $I \oplus I \not\simeq R \oplus I^2$.

Poof. Because $I$ is not invertible, it follows from [1, Proposition 20.8], that $\text{Fitt}_1(I) \neq R$. By the above Proposition, we infer that $\text{Fitt}_3(I \oplus I) = \text{Fitt}_1(I) \neq R = \text{Fitt}_3(R \oplus I^2)$.

For a concrete instance, we can set $$R = \mathbb{Z} + 2i \mathbb{Z}, \, I = 2\mathbb{Z}[i].$$ It is easily checked that $I^{-1} = \mathbb{Z}[i]$ so that $I$ is not invertible and $\text{Fitt}_1(I) = II^{-1} = I$. As $I \simeq I^2$, we have in addition $\text{Fitt}_2(R \oplus I^2) = \text{Fitt}_1( I^2) = \text{Fitt}_1(I) = I$ whereas $\text{Fitt}_2(I \oplus I) = I^2 \neq I$ because of the above Proposition.

We conclude with a general positive result relating direct sum of ideals to products.

Theorem [2, C1]. Let $R$ be an integral domain and let $I_1, \dots, I_r$ be ideals of $R$. Set $M = I_1 \oplus \cdots \oplus I_r$ and let $T$ be the torsion submodule of $\Lambda^r(M)$. Then $\Lambda^r(M) /T \simeq I_1 \cdots I_r$. Therefore if $J_1, \dots, J_r$ are ideals of $R$ such that $I_1 \oplus \cdots \oplus I_r \simeq J_1 \oplus \cdots \oplus J_r$, then $I_1 \cdots I_r \simeq J_1 \cdots J_r$.


  • [1] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 2004.
  • [2] H. Matsumura, "Commutative Ring Theory", 1989.
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